The length $\ell$ of a rectangle is decreasing at a rate of $2 \mathrm{~cm} / \mathrm{sec}$ while the width $w$ is increasing at a rate of $2 \mathrm{~cm} / \mathrm{sec}$. When $\ell=3 \mathrm{~cm}$ and $\mathrm{w}=4 \mathrm{~cm}$, find the rates of change of the area, the perimeter, and the lengths of the diagonals of the rectangle. Determine which of these quantities are increasing, decreasing, or constant.

Step 1 ：The area $A$ of a rectangle is given by $A = \ell w$, the perimeter $P$ is given by $P = 2(\ell + w)$, and the length $d$ of the diagonal is given by Pythagoras' theorem $d = \sqrt{\ell^2 + w^2}$.

Step 2 ：We are given that $\frac{d\ell}{dt} = -2$ and $\frac{dw}{dt} = 2$, and we are asked to find $\frac{dA}{dt}$, $\frac{dP}{dt}$, and $\frac{dd}{dt}$ when $\ell = 3$ and $w = 4$.

Step 3 ：We can find these by differentiating the expressions for $A$, $P$, and $d$ with respect to time $t$ and then substituting the given values.

Step 4 ：By differentiating and substituting the given values, we find that $\frac{dA}{dt} = -2$, $\frac{dP}{dt} = 0$, and $\frac{dd}{dt} = \frac{2}{5}$.

Step 5 ：The area is decreasing at a rate of $2 \mathrm{~cm}^2 / \mathrm{sec}$, the perimeter is not changing, and the length of the diagonal is increasing at a rate of $\frac{2}{5} \mathrm{~cm} / \mathrm{sec}$.

Step 6 ：Final Answer: The rates of change of the area, the perimeter, and the lengths of the diagonals of the rectangle when $\ell = 3$ and $w = 4$ are $\boxed{-2 \mathrm{~cm}^2 / \mathrm{sec}}$, $\boxed{0 \mathrm{~cm} / \mathrm{sec}}$, and $\boxed{\frac{2}{5} \mathrm{~cm} / \mathrm{sec}}$, respectively.